Post on 16-Mar-2020
Hard
1. A car is stuck in the mud. The mud splashes from the rim of a rotating tire of radius π spinning at
a speed π£ where π£2 > ππ . Neglect the resistance of the air. What is the maximum height the
mud can rise above the ground?
Solution:
Energy conservation ππβ0 =ππ£2(cos πΌ)2
2. The total height is π» = π + π π ππ πΌ +
π£2(cos πΌ)2
2π .
Minimizing this under assumption π£2 > ππ gives sin πΌ0 =ππ
π£2
We obtain π» = π +π£2
2π+
ππ 2
2π£2
Easy
1. A capacitor is made from two hollow, coaxial copper cylinders, one inside the other. There is air
in the space between the cylinders. The inner cylinder has net positive charge and the outer
cylinder has net negative charge. The inner cylinder has radius π1 = 2.30 mm , the outer cylinder
has radius π2 = 2.90 mm , and the length of each cylinder is L=36.0 cm.
A. If the potential difference between the surfaces of the two cylinders is 60.0 V , what is the
magnitude of the electric field at a point between the two cylinders that is a distance of 2.60
mm from their common axis and midway between the ends of the cylinders?
B. What is the capacitance of this capacitor?
Solution:
From Gaussβs law πΈ = π
πΏ 2ππ π0 . Integrating we can find the potential difference π =
π
πΏ2ππ0 ln
π2
π1 .
The answer for A. becomes πΈ =π
πππ(π2π1
)
The answer for B. becomes πΆ =π
π=
πΏ2ππ0
lnπ2π1
2. A parallel-plate capacitor has square plates that are π = 5.00 cm on each side and π = 3.50 mm
apart. The space between the plates is completely filled with two square slabs of dielectric, each
5.00 cm on a side and π1 = 2.00, π2 = 1.50 mm thick. One slab is pyrex glass and the other is
polystyrene.
If the potential difference between the plates is π = 90.0 V , how much electrical energy is
stored in the capacitor?
Solution:
The problem is equivalent to having two capacitors in series. π =πΆπ2
2=
π
2
11
πΆ1+
1
πΆ2
where πΆπ =π0π2
πππ π
EM-easy-1
A circular coil 0.0500 meter in radius, with 30 turns of wire, lies in a horizontal plane. It carries a
current of 5.00 A in a counterclockwise sense when viewed from above. The coil is in a uniform
magnetic field directed toward the right, with magnitude 1.20 T.
Find
(a) The magnitudes of the magnetic moment
(b) The torque on the coil.
(c) The magnitudes of force on the coil
The force is zero
EM-easy-2
A bridge circuit is shown in the Figure.
(a) Find the current in each resistors and the equivalent resistance of the network of five
resistors
(b) Find the potential difference Vab (Vab= Va βVb)
EM-easy-3 (could be used for EM-hard)
A rod with length L, mass m, resistance R on a fixed rail on a horizontal work station under a
uniform magnetic field. It starts to move with an initial velocity of vo, as shown in the Fig.
(Ignore friction and resistance of the rail).
How far will the rod go?
EM-hard-1
A capacitor is made of two concentric cylinders of radius r1 and r2 (r1 < r2) and length L >> r2.
The region between r1 and r3 = (r1* r2)1/2 is filled with a circular cylinder of length L and
dielectric constant K (the remaining volume is an air gap).
What is the capacitance?
EM-hard-2
Two semi-infinite conductive plates A and B are grounded and connected at O, with a point
charge Q placed in the acute angle between two plates, as shown below. The distance from point
O to charge Q is R. Find the electric potential in the space between the two plates.
ME-hard-1
Let us consider the motion of a three-particle system in which all the particles lie in a straight
line, such as the carbon dioxide molecule CO2. We consider motion only in one dimension, along
the x-axis (as shown in the Fig.). The two end particles, each of mass m, are bound to the central
particle, mass M, via a potential function that is equivalent to that of two spring of stiffness K, as
shown in Fig. The coordinated expressing the displacements of each mass are x1, x2, and x3.
Find
(a) The Lagrangian of the system
(b) The three equations of motion
(c) The normal mode frequencies
Hard
1. An ac parallel RLC circuit consists of a voltage source of frequency π, a resistor π , a capacitor
of capacitance πΆ, and inductor πΏ.
A. What must the current amplitude πΌ through the inductor be for the average electrical
power consumed in the resistor to be π?
B. Draw the energy dissipation in R, L and C as a function of time.
C. Calculate the power factor.
Solution:
We have ππΏ = ππΏ. The current amplitude πΌ = π0/ππΏ . We also have π =π0
2
2π for the resistor -> πΌ =
β2π π
ππΏ.
The power factor= π
π02/2π
=π
π =
1
β1+(π ππΆβπ
ππΏ)
2
2. Two square metal plates of side πΏ are separated by a distance π much smaller than πΏ. A
dielectric slab of size πΏ Γ πΏ Γ π slides between the plates. It is inserted at a distance π₯
(parallel to one side of the squares) and held there (see Figure)
A. Find the force exerted electrically on the slab. Be careful and explicit about its direction.
B. How does the situation change if the battery is left connected?
Solution:
The energy is given by π =π2
2πΆ=
πΆπ2
2 . The capacitance πΆ =
π0πΏ
π(π π₯ + πΏ β π₯)
A. πΉ = βππ
ππ₯=
π0πΏπ2
2π(π β 1) Drawn further between the plates.
B. πΉ = βΞ(
πΆπ2
2)βπΞπ
Ξπ₯=
π0πΏπ2
2π(π β 1)
d
x